Question

\(A\) thin plate is bounded by the graphs of \(y=e^{-x}, y=-e^{-x}, x=0,\) and \(x=L .\) Find its center of mass. How does the center of mass change as \(L \rightarrow \infty ?\)

Short answer

Based on the solution above, we can conclude that as the width of the thin plate (L) approaches infinity, the center of mass moves infinitely far to the right and gets closer to the x-axis. In other words, the x-coordinate of the center of mass approaches infinity, and the y-coordinate approaches zero.

Step-by-step solution

Find the area of the thin plate

The area of the thin plate can be found by subtracting the lower function (the negative function) from the upper function (the positive function) and integrating this difference in the given interval. For example, we can find the area of the thin plate (A) as follows: \(A = \int_0^L (e^{-x} - (-e^{-x}))dx = \int_0^L 2e^{-x} dx\) Now, we will evaluate this integral to find the area. \(A = \left[-2e^{-x}\right]_0^L = 2[1 - e^{-x}]\)

Find the mass of the thin plate

Next, we will find the mass of the thin plate. Assume the density of the material is \(\rho\). Since the density is constant, we can find the mass (M) by multiplying the area (A) by the density: \(M = \rho A = \rho(2[1 - e^{-L}])\)

Find the coordinates of the center of mass

Now we will find the coordinates of the center of mass. \(x\) and \(y\) coordinates of the center of mass can be computed using the following formulas: \(x_{CM} = \frac{1}{M} \int_0^L x \times(age^{-x}+(-e^{-x}))dx\) \(y_{CM} = \frac{1}{M} \int_0^L y \times(2e^{-x}dy)\) First, we will determine \(x_{CM}\): \(x_{CM} = \frac{1}{\rho (2[1 - e^{-L}])} \int_0^L x(2e^{-x})dx\) To evaluate this integral, use integration by parts: Let \(u = x, \; du = dx\) Let \(dv = 2e^{-x}dx, \; v = -2e^{-x}\) \(x_{CM} = \frac{1}{\rho(2[1-e^{-L}])} \left[-2xe^{-x} \right]_0^L + \frac{1}{\rho(2[1-e^{-L}])} \int_0^L 2e^{-x} dx\) \(x_{CM} = \frac{1}{\rho(2[1-e^{-L}])} \left[-2Le^{-L}\right] + \frac{1}{\rho(2[1-e^{-L}])} (2[1 - e^{-L}])\) \(x_{CM} = \frac{-2Le^{-L} + 2(1 - e^{-L})}{2[1-e^{-L}]}\) Now we'll find \(y_{CM}\): \(y_{CM} = \frac{1}{\rho(2[1 - e^{-L}])} \int_0^L y(2e^{-x}dy)\) To evaluate this integral, we'll first change the variable. Let \(y = e^{-x}\), so \(dy = -e^{-x} dx\) Now use the limits of integration \(y = e^{-L}\) and \(y = 1\): \(y_{CM} = \frac{1}{\rho(2[1 - e^{-L}])} \int_{e^{-L}}^1 2y dy\) \(y_{CM} = \frac{1}{\rho(2[1 - e^{-L}])} \left[y^2\right]_{e^{-L}}^1\) \(y_{CM} = \frac{1}{\rho(2[1 - e^{-L}])} \left[1 - e^{-2L}\right]\) Now we have both coordinates of the center of mass.

Investigate the behavior of the center of mass as L approaches infinity

Finally, let's analyze how the center of mass changes as L approaches infinity. As \(L\to\infty\), it is clear that the denominator of \(x_{CM}\) approaches zero: \(x_{CM} = \frac{-2Le^{-L} + 2(1 - e^{-L})}{2[1-e^{-L}]}\) As the denominator approaches zero, \(x_{CM}\) approaches infinity. This means that as the thin plate becomes infinitely wide, its center of mass moves infinitely far to the right. Similarly, for \(y_{CM}\): \(y_{CM} = \frac{1}{\rho(2[1 - e^{-L}])} \left[1 - e^{-2L}\right]\) As \(L\to\infty\), the denominator approaches infinity, and the numerator approaches 1: \(y_{CM} = \frac{1 - e^{-2L}}{2[1 - e^{-L}]}\) As the denominator approaches infinity, \(y_{CM}\) approaches zero. This means that as the thin plate becomes infinitely wide, its center of mass moves closer to the x-axis.

Key concepts

Integration

Integration is a fundamental concept in calculus that is used to find areas under curves, among other applications. For this exercise, we use integration to determine the area of the region under the curve, which helps us in calculating both the area and mass of the plate. Consider the area between two functions, here specifically between the curves of the exponential functions:

• Area (\(A\)) is calculated using: \[ A = \int_0^L \left( e^{-x} - (-e^{-x}) \right) dx = \int_0^L 2e^{-x} dx \]By evaluating this integral, we determines the area which is pivotal for center of mass calculations.
• The result after integration helps establish a foundation for determining the mass, as mass is the product of density and area.

Without mastering integration, dealing with continuous data would be much more challenging.

Density

Density is a crucial concept in physics and calculus when discussing mass distribution. It refers to the amount of mass per unit area or volume. In this context, it is used to calculate the mass of the thin plate which is central to finding the center of mass. Here's how it plays a role:

• Density (\(\rho\)) is assumed constant across the plate.
• With the area \(A = 2[1 - e^{-L}]\) known, the mass (\(M\)) of the plate is determined by \(M = \rho A\), to yield \(M = \rho(2[1 - e^{-L}])\).
• This highlights how density ties together geometric data (area) and physical properties (mass).

By understanding density, you gain insight into how material properties affect overall weight and balance, crucial for real-world applications.

Limits at Infinity

When discussing limits in calculus, understanding behavior "at infinity" is invaluable, especially for evaluating how functions behave as inputs grow large. Here, when the length \(L\) of the thin plate extends towards infinity, we analyze:

• The \(x_{CM}\) of the center of mass calculation: \[x_{CM} = \frac{-2Le^{-L} + 2(1 - e^{-L})}{2[1-e^{-L}]}\] As \(L \to \infty\), the term \(xe^{-x}\) dwindles, propelling \(x_{CM}\) even further to infinity, showing how \(x_{CM}\) moves infinitely far to the right.
• For \(y_{CM}\): \[y_{CM} = \frac{1 - e^{-2L}}{2[1 - e^{-L}]}\] The numerator approaches 1 while the denominator becomes infinite, making \(y_{CM}\) asymptotically approach zero.

This analysis of limits allows us to predict the behavior of mathematical models as parameters transform towards extreme values, such as here, when the plate width increases without bound.